SolveValues[x^3 == -1, x]? (Returns {-1, (-1)^(1/3), -(-1)^(2/3)}.)

Addendum

Also: Exp[Log[-1] + Range[3] 2 Pi/3 I] // Simplify

The warning seems to apply to the value of the parameter y -> 1.

The issue is that the complex-valued principal root does not have -1 in its range. Consider

Arg[x^(1/3)]
(* Arg[x] / 3 *)

It implies that the argument of x^(1/3) is between ±Pi/3, since Arg[x] is between ±Pi. There are three cube roots of any nonzero complex number, and their arguments differ by 2Pi/3. There is always exactly one in the sector -Pi/3 < Arg[y] <= Pi/3. This root is what x^(1/3) represents.

Therefore the real part of x^(1/3) is greater than or equal to zero. I believe that Solve[] is looking at all roots in the complex plane, but there are none.

This also works:

In[10]:= Solve[CubeRoot[x] == -1, x]

Out[10]= {{x -> -1}}

However, it seems to me that Solve should look at all the roots in the complex plane.

Thanks a lot. I feel that these things should be emphasized in learning Math and refines the concept of roots.

Which is true according to a rigorous mathematical definition？

Sorry, I meant x^(1/3)=Exp[Log[x]/3].

And why

SolveValues[x^3 == -1, x] // ComplexExpand

gives three answers:

{-1, 1/2 + (I Sqrt[3])/2, 1/2 - (I Sqrt[3])/2}

So, what is/are the exact solution/solutions for

x^3=-1

Hello, May you teach me why

(-1)^(1/3) // N

gives result containing i?

As

a=b

of course

a^3=b^3

Where am I wrong?

SolveValues[x^3 == -1, x] and Solve[x^3 == -1, x] give me the same solutions (in different forms). Or maybe that's what you meant. I wasn't sure.

-1 has three cube roots.

In[19]:= {-1, Exp[I \[Pi]/3], Exp[-I \[Pi]/3]}^3

Out[19]= {-1, -1, -1}

How can I compute all 3 of them in Mathematica?